The 11-cell or hendecachoron is 11 hemi-icosahedra. I have a hemi-icosahedra (OFF file below). I was wondering if the 11-cell might be attainable (in Stella4D, pressing "u") but it doesn't appear to be. It stops at 10 cells and can't display anything. Is this because the hemi-icosahedra is non-orientable?

Being non-orientable doesn't matter (this is already the case for many uniform polychora, is it not?).

I imagine the problem is that coplanar faces of the vertex figure share an edge. Maybe the fact that those coplanar faces also pass through the centre of the model makes things worse. Actually I'd be surprised if that combo did work.

robertw wrote:Maybe the fact that those coplanar faces also pass through the centre of the model makes things worse.
Rob.

Carlo's models always were tweaked so that the intersection didn't happen. I can try that.

I remember some of profesional geometers saying that the 11-cell and 57-cell can't be represented in the same way as other polytopes. Something to do with hyberbolic geometry; I have little hope of understanding it.

I made the network for the hemi-dodecahedron. There isn't much promise that this figure can be depicted in anyway to construct the 57-cell.

I don't really understand the shape you're after. Hitting "u" is for uniform or scaliform polychora. Does the 11-cell have all vertex figures the same? I can't imagine that adjusting it to avoid the coplanarity could possibly work. If as you mention that it can only be realised in hyperbolic space, then you won't be able to create it in Stella4D.

Carlos' paper is on line. It shows ways of trying to display the 11-cell but using purely 3D constructs. i.e usings 11 vertices and so on. I was bring such models into Stella but I had to double the number of faces (replicated on top of one another) in order for it to be a closed system. This also triples the number of edges. However, since these faces and edges are visually simultaneous, it is still a representation.

I'm aware of two potential hendecachora with identical cells that could be modeled in 4D, however they aren't the one with the hemi-icosahedra - that one requires some strange curved 4-space. I have yet to model the two 11-celled polychora, but I have done the 7-cell, two 9-cells, and three distinct 13-celled polychora. These polychora belong to a group that Wendy Kreiger calls "step-tegums". Below is the net of one of the 13-cells.

Jabe wrote:I'm aware of two potential hendecachora with identical cells that could be modeled in 4D, however they aren't the one with the hemi-icosahedra - that one requires some strange curved 4-space. I have yet to model the two 11-celled polychora, but I have done the 7-cell, two 9-cells, and three distinct 13-celled polychora. These polychora belong to a group that Wendy Kreiger calls "step-tegums". Below is the net of one of the 13-cells.

Perhaps those would be nice additions to the Stella library. Are the 7 though 13 (odd) cells unique or are there N-cells at various integer values?

I suppose the hemi-icosahedron has to be modeled in a curved 4-space if all the faces are to be the same. I found something interesting, however, if it means anything significant I am not sure.

The connection system for the hemi-icsoahedron can be made in 3D, and I suppose it could be shown to be mobius but doing so is very convoluted. If it means one could hop from face to face and eventually be on the other side of the starting face that would be simple. But from each face there would be two paths to take. The total number of circuits would be fairly high I presume.

Let it be such, though, that it can be represented in 3-space with more than one face type, it can also be represented in 2-space. Smashed flat, it will still have the 6 vertices, 10 faces, and 15 edges. Those 6 vertices can be placed on the 6 of the 11 vertices of a heptagon. This can be done in Stella.

Using an 11-gonal prism, 11.4.4, one one face count off the vertices 1 through 11. Change to No Symmetry, facet out 10 triangles between these vertices.

1 2 5
1 2 8
1 3 6
1 3 8
1 5 6
2 3 5
2 3 6
2 6 8
3 5 8
5 6 8

This closes and creates the flat semi-icosahedron. Now change to 11-fold Pyramidal symmetry. This repeats the former faceting 11 times around the heptagon. From the faceting window save the faceting diagram as an OFF file (nice trick in Stella!).

Looking at the OFF file in an editor, it has 11 vertices, 55 faces, and 55 edges, the numbers corresponding to the 11-cell. Apparently, for every hemi-icosahedron 5 of the faces are ajoined to 5 others cancelling them out.

Unfortunately as is, the OFF file can't be brought back into Stella. Being flat is bad enough, one can tweak one of the Z-axis coordinates but that is not enough. The faces need to be doubled to get two faces to an edge, then it can be brought into Stella, then it may become the compound of 10 hemi-icosahedra.

Something fun to try if interested. I don't think is the only configuration of hemi-icosahedron that can do this. Simulating the 57-cell on the 57-gon might be possible.

Makes me wonder if an icosahedron could be smashed flat and repeated on some kind of polygon.

polymorphic wrote:And the number of configurations increases as N grows larger I would presume. Is there a formula for that?

Roger

A step tegum requires two numbers - N and M - where N is the number of cells, and M represents the "step". I'll use a 13 sided one for an example.

Consider the 13-gon, we can label its vertices from 0 to 12 - now consider the 13-gon duoprism, we can label its vertices with two numbers - both going from 0 to 12. To find the vertices of the 13 - 3 step prism (where 3 is the step), we start with vertex 0,0 - then 1,3 - 2,6 - 3,9 - 4,12 - 5,2 (2 is 15 mod 13) - 6,5 - 7,8 - 8,11 - 9,1 - 10,4 - 11,7 - 12,10. The step prism is the convex hull of those points, where the step tegum is the dual.

So the question is "what M's give us distinct step tegums" - when M = 1, we only get a polygon - so it never works. When M > N/2 it looks just like a N-N/2 step tegum (in otherwords, it might be a step 10 if we count the vertices clockwise, but going counterclockwise it would be a step 3 - so the only M's to check are those between 2 and N/2. N/2 gives us a duotegum (dual of duoprism) - so no need to count it. So for 13, M can be 2,3,4,5, or 6 - but there is one more trick - a 13-gon duoprism is the cross product of 13-gon #1 by 13-gon #2 - if we change the order to 13-gon #2 by 13-gon #1, the step may also change - it turns out that 13 - 2 = 13 - 6, what is step 2 in one orientation is step 6 in the other. Here is a trick to see which steps are equal: Let M and P be two different steps, if MP (M times P) is one off from a multiple of N, then N-M = N-P. 2x6 = 12 which is one off from 13, 3x4 = 12 which is one off from 13, the only number left is 5 - 5x5 = 25 is one off from 26 a multiple of 13. The 13-5 has twice the symmetry also. We can call P the alter-step of M

One more thing, if N and M are not relatively prime - then M will not pair up with another step - since MP is never one off from a multiple of N. So there are three types of step tegums: M,N not relatively prime (i.e. 10-4), M,N relatively prime and M<>P (P is the alter-step), and M,N relatively prime with M=P (has double symmetry).

It looks like there is no other way to make these than by hand. It is amazing to me that such an oddball looking outer cell can contain copies of itself inside.

Do you have the stel file for this? I had thought I might try to construct what you had just by looking but I'd have to guess at too much. But you must have to do this by hand, so I am wondering how you do it.